Solve 3 variables with 2 equations

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Since one can assume that the full price is greater than $3.10, the full price expressed in cents is greater than 310. To solve the problem, we need to find a pair of integers x and y such thatx + y = 1653andx + 225y = 512500given that both x and y are positive integers and that x is greater than 310. There are a finite number of such pairs, so it is an easy matter to find the solution. However, I do not know an easy way to find the solution. The solution is x = 1348 and y = 305. Therefore, 305 items were sold at the sale price of $2.25.
  • #1
abenedet
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Here is the problem:

Total sales are $5125 and units sold are 1653. Units can be sold either on sale or not on sale. The price on sale or feature sale price (FSP) was $2.25. How many units were sold on sale?

I can derive 2 formulas, but I have 3 variables to solve. Here's how far I have gotten:

Regular price = RSP
Units sold on Sale = Promoted volume (PV)
Units sold at regular = non-promoted volume (NPV)
Total volume = TV

Equation 1: TV = PV + NPV
Equation 2: Total Sales = (RSP x NPV) + (FSP x PV) or $5125 = (RSP x NPV) + 2.25PV

I cannot determine another equation to help solve this. Any help will be greatly appreciated.
 
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  • #2
abenedet said:
Here is the problem:

Total sales are $5125 and units sold are 1653. Units can be sold either on sale or not on sale. The price on sale or feature sale price (FSP) was $2.25. How many units were sold on sale?

I can derive 2 formulas, but I have 3 variables to solve. Here's how far I have gotten:

Regular price = RSP
Units sold on Sale = Promoted volume (PV)
Units sold at regular = non-promoted volume (NPV)
Total volume = TV

Equation 1: TV = PV + NPV
Equation 2: Total Sales = (RSP x NPV) + (FSP x PV) or $5125 = (RSP x NPV) + 2.25PV

I cannot determine another equation to help solve this. Any help will be greatly appreciated.

Homework Statement

Let's see: you have RSP, PV, NPV, TV, and Total Sales. I think you have way too many variables that aren't important and not enough of the variables you are actually looking for. Plus, there is some information given that you didn't use. The way I read this is an item sells for its normal price or its sale price. What are they looking for in this problem?

Hint: One equation should give the total number of items, and the other should give the total revenue for the two types of items (regular price items and on-sale items).
 
  • #3
Mark44 said:
Let's see: you have RSP, PV, NPV, TV, and Total Sales. I think you have way too many variables that aren't important and not enough of the variables you are actually looking for. Plus, there is some information given that you didn't use. The way I read this is an item sells for its normal price or its sale price. What are they looking for in this problem?

They are looking for how many items were sold at the sale price. Also, the only known variables I have are total volume, FSP, and total Sales.

Mark44 said:
Hint: One equation should give the total number of items, and the other should give the total revenue for the two types of items (regular price items and on-sale items).

Yes, these are the equations that I was able to derive, but I then got stuck because I have 3 variables to solve in these two equations in order to derive how many items were sold at sale price:

Equation 1: TV = PV + NPV
Equation 2: Total Sales = (RSP x NPV) + (FSP x PV) or $5125 = (RSP x NPV) + 2.25PV

Given these two equations (or any other equations that you can think of), how can I solve for PV?
 
  • #4
abenedet said:
They are looking for how many items were sold at the sale price. Also, the only known variables I have are total volume, FSP, and total Sales.
If a number is known, it's not a variable and shouldn't be represented as a variable.

For example, total volume is known, total sales are known, and sale price is known.

After eliminating all "variables" that aren't actually variable, I get two equations in three unknowns, as well. Are you sure you have provided all the information in this problem?
abenedet said:
Yes, these are the equations that I was able to derive, but I then got stuck because I have 3 variables to solve in these two equations in order to derive how many items were sold at sale price:

Equation 1: TV = PV + NPV
Equation 2: Total Sales = (RSP x NPV) + (FSP x PV) or $5125 = (RSP x NPV) + 2.25PV

Given these two equations (or any other equations that you can think of), how can I solve for PV?
 
  • #5
Yes, you are correct...my terminology was misleading. There are 3 unknowns. Sorry for the confusion.

And yes, this is all the information that I have.

Thanks
 
  • #6
I used variables of x, y, and FP.
x = # of items sold at full price
y = # of items sold at sale price of $2.25

Altogether 1653 items were sold, so it's possible to get one equation in two variables from this information. From this equation, you can solve for y, the number of items sold at the sale price, in terms of the full price.

It's reasonable to assume that both x and y are greater than 0, and that both x and y have integer values. This means that the full price has to be larger than $3.10, which yields a negative number of items sold.
 
  • #7
One solution to the problem (there are other solutions) is arrived at by assuming that all but one item were sold at the sale price. This would mean the full price is $1408, a price that makes the variations in the cost of an airplane seat look mild.

If one assumes that the full price expressed in cents is a positive integer, the problem becomes a fairly simple second order Diophantine problem.
 

1. How do you solve for 3 variables with 2 equations?

To solve for 3 variables with 2 equations, you will need to use a method called substitution. This involves isolating one variable in one of the equations and substituting its value into the other equation. Then, you can solve for the remaining two variables. Repeat this process for the other two variables to find their values.

2. Can you solve for 3 variables with 2 equations if they are not linear?

No, you cannot solve for 3 variables with 2 equations if they are not linear. This is because substitution only works for linear equations, where the variables have a power of 1. If the equations are not linear, you will need to use other methods such as elimination or graphing.

3. Is it possible to have more than one solution when solving for 3 variables with 2 equations?

Yes, it is possible to have more than one solution when solving for 3 variables with 2 equations. This usually happens when the equations are dependent, meaning they represent the same line or plane. In this case, any values that satisfy one equation will also satisfy the other, resulting in infinitely many solutions.

4. Can you solve for 3 variables with 2 equations if the equations are not independent?

No, you cannot solve for 3 variables with 2 equations if the equations are not independent. This is because independent equations have a unique solution, while dependent equations have infinitely many solutions. If the equations are not independent, it means they are dependent, and you will not be able to solve for all 3 variables.

5. Are there any real-life applications for solving 3 variables with 2 equations?

Yes, there are many real-life applications for solving 3 variables with 2 equations. This method is commonly used in economics to determine the optimal combination of goods and services to produce, taking into account limited resources and market demand. It is also used in engineering to solve systems of equations representing physical phenomena, such as electrical circuits or chemical reactions.

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